Question 5.1: For the flat steel spring shown in Figure 5–6, compute the m...

Objective: Compute the maximum, minimum, mean, and alternating tensile stresses in the flat spring. Compute the stress ratio, R.

Given: Layout shown in Figure 5–6. The spring is steel: L = 65 mm.
Spring cross-sectional dimensions: t = 0.80 mm and b = 6.0 mm.
Maximum deflection of the spring at the follower = 8.0 mm.
Minimum deflection of the spring at the follower = 3.0 mm.

Analysis: Point A at the base of the spring experiences the maximum tensile stress. Determine the force exerted on the spring by the follower for each level of deflection using the formulas from Table A14–2, Case (a).
Compute the bending moment at the base of the spring for each deflection. Then compute the stresses at point A using the bending stress formula, \sigma=M c / I. Use Equations (5–1) (\sigma_{m}=\left(\sigma_{\max }+\sigma_{\min }\right) / 2), to (5–3) (\text { Stress ratio } R=\frac{\text { minimum stress }}{\text { maximum stress }}=\frac{\sigma_{\min }}{\sigma_{\max }}, \text { Stress ratio } A=\frac{\text { alternating stress }}{\text { mean stress }}=\frac{\sigma_{a}}{\sigma_{m}}) for computing the mean, alternating stresses, and R.

Results: Case (a) of Table A14–2 gives the following formula for the amount of deflection of a cantilever for a given applied force:

y=P L^{3} / 3 E I
Solve for the force as a function of deflection:
P=3 E I y / L^{3}

Appendix 3 gives the modulus of elasticity for steel to be E = 207 GPa. The moment of inertia, I, for the spring cross section is found from

I=b t^{3} / 12=(6.00 \mathrm{~mm})(0.80 \mathrm{~mm})^{3} / 12=0.256 \mathrm{~mm}^{4}
Then the force on the spring when the deflection y is 3.0 \mathrm{~mm} is
P=\frac{3\left(207 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.256 \mathrm{~mm}^{4}\right)(3.0 \mathrm{~mm})}{(65 \mathrm{~mm})^{3}} \frac{\left(1.0 \mathrm{~m}^{2}\right)}{\left(10^{6} \mathrm{~mm}^{2}\right)}=1.74 \mathrm{~N}
The bending moment at the support is
M=P \cdot L=(1.74 \mathrm{~N})(65 \mathrm{~mm})=113 \mathrm{~N} \cdot \mathrm{mm}
The bending stress at point A caused by this moment is
\sigma=\frac{M c}{I}=\frac{(113 \mathrm{~N} \cdot \mathrm{mm})(0.40 \mathrm{~mm})}{0.256 \mathrm{~mm}^{4}}=176 \mathrm{~N} / \mathrm{mm}^{2}=176 \mathrm{MPa}

This is the lowest stress that the spring sees in service, and therefore \sigma_{\min }=176 \mathrm{MPa}.
Because the force on the spring is proportional to the deflection, the force exerted when the deflection is 8.0 \mathrm{~mm} is
P=(1.74 \mathrm{~N})(8.0 \mathrm{~mm}) /(3.0 \mathrm{~mm})=4.63 \mathrm{~N}
The bending moment is
M=P \cdot L=(4.63 \mathrm{~N})(65 \mathrm{~mm})=301 \mathrm{~N} \cdot \mathrm{mm}
The bending stress at point A is
\sigma=\frac{M c}{I}=\frac{(301 \mathrm{~N} \cdot \mathrm{mm})(0.40 \mathrm{~mm})}{0.256 \mathrm{~mm}^{4}}=470 \mathrm{~N} / \mathrm{mm}^{2}=470 \mathrm{MPa}
This is the maximum stress that the spring sees, and therefore \sigma_{\max }=470 \mathrm{MPa}.
Now the mean stress can be computed:
\sigma_{m}=\left(\sigma_{\max }+\sigma_{\min }\right) / 2=(470+176) / 2=323 \mathrm{MPa}
Finally, the alternating stress is
\sigma_{a}=\left(\sigma_{\max }-\sigma_{\min }\right) / 2=(470-176) / 2=147 \mathrm{MPa}
The stress ratio is found using Equation (5-3):
Stress ratio R=\frac{\text { minimum stress }}{\text { maximum stress }}=\frac{\sigma_{\min }}{\sigma_{\max }}=\frac{176 \mathrm{MPa}}{470 \mathrm{MPa}}=0.37

Comments: The sketch of stress versus time shown in Figure 5–4(a) illustrates the form of the fluctuating stress on the spring. In Section 5–8, you will see how to design parts subjected to this kind of stress.

TABLE A14–2 Beam-Deflection Formulas for Cantilevers
At B at end: y_{B}=y_{\max }=\frac{-P L^{3}}{3 E I} Between A and B: y=\frac{-P x^{2}}{6 E I}(3 L-x)
At B at load: y_{B}=\frac{-P a^{3}}{3 E I} At C at end: y_{C}=y_{\max }=\frac{-P a^{2}}{6 E I}(3 L-a) Between A and B : y=\frac{-P x^{2}}{6 E I}(3 a-x) Between B and C : y=\frac{-P a^{2}}{6 E I}(3 x-a)
W=\text { total load }=w L At B at end: y_{B}=y_{\max }=\frac{-W L^{3}}{8 E I} Between A and B : y=\frac{-W x^{2}}{24 E I L}\left[2 L^{2}+(2 L-x)^{2}\right]
M_{B}= concentrated moment at end At B at end: y_{B}=y_{\max }=\frac{-M_{B} L^{2}}{2 E I} Between A and B : y=\frac{-M_{B} x^{2}}{2 E I}